Question

The general solution of the differential equation $(x+y+3)\frac{dy}{dx}=1$ is

• A. $x+y+3=C{e}^y$
• B. $x+y+4=C{e}^y$
• C. $x+y+3=C{e}^{-y}$
• D. $x+y+4=C{e}^{-y}$
• E. $x+y+4e^y=C$

Answer 1

Answer: (B)

We have, $(x+y+3)\frac{dy}{dx}=1$
$\Rightarrow (x+y+3)=\frac{dx}{dy}$
Let $x+y+3=t$
On differentiating w.r.t.y, we get
$\frac{dx}{dy}+1=\frac{dt}{dy}$
$\Rightarrow \frac{dt}{dy}=t+1\left[\because \text{ from Eq. (i), }t=\frac{dx}{dy}\right]$
On integrating both sides,
$\int \frac{dt}{(t+1)}=\int dy$
$\Rightarrow \log (t+1)=y+C_1$
$\Rightarrow \log (x+y+3+1)=y+C_1$
$\therefore x+y+4=C{e}^y$ [where ,$c=e^{c_1}$]